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3 years ago

The equation of a circle centered at the origin is x^2+y^2=16. what is the radius of the circle?

Mathematics

1 answer:

vekshin13 years ago

4 0

Answer:

The center is (0,0) and the radius is 4

Step-by-step explanation:

x^2+y^2=16.

The equation of a circle can be written in the form

(x-h)^2+(y-k)^2=r^2 where ( h,k) is the center and r is the radius

(x-0)^2+(y-0)^2=4^2

The center is (0,0) and the radius is 4

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5 less than the product of 8 and a number

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3 years ago

Read 2 more answers

Assume that there are an equal number of births in each month so that the probability is that a person chosen at random was born

NikAS [45]

Answer:

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they have a birthday in May, or they do not. The probability of a person having a birthday in May is independent of any other person. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of a person being in May:

May has 31 days in a year of 365. So

$p = \frac{31}{365} = 0.0849$

Group of 20 friends:

This means that $n = 20$

What is the probability that at the May celebration, exactly two members of the group have May birthdays?

This is P(X = 2).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{20,2}.(0.0849)^{2}.(0.9151)^{18} = 0.2773$

0.2773 = 27.73% probability that at the May celebration, exactly two members of the group have May birthdays

3 0

3 years ago

A deck of cards contains RED cards numbered 1,2,3 and BLUE cards numbered 1,2,3,4,5,6. Let R be the event of drawing a red card,

Tatiana [17]

Answer:

e. R′

Step-by-step explanation:

Let the Sample Space be a universal set consisting of 3 red and 6 blue cards. Then the event of of getting 4 blue cards will be given

R complement = R` = Universal Set minus Red cards will give blue cards.

a. R OR O

It cannot be this option because we need 4 blue card

b. B AND O

It cannot be this choice as well because 4 is not odd.

c. R OR E

4 is even but we need blue

d. R AND O

Red and odd is again not required

e. R′ = It will give our required result.

f. E′= 4 is even if it is complemented it cannot be obtained and will be left out.

5 0

4 years ago